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3 years ago

6

How do I do this? I dont know how to work out his problem

1 answer:

mrs_skeptik [129]3 years ago

8 0

The mom is 36

The twins are 11

1. 58-25= 33

2. 33/3= 11 <----The twins age

3. 11+25= 36 <----The moms age

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Type the correct answer in the box.

natita [175]

Answer:

250 cm^3.

Step-by-step explanation:

Volume of the prism

= 12 * 5^2

= 300 cm^3.

Volume of the pyramid

= 1/3 * 25 * 6

= 50 cm^3

The volume of the space

= 300 - 50

250 cm^3.

4 0

3 years ago

The median of the values in a data set is x. If 32 were subtracted from each

Orlov [11]

Answer:

The answer is B: (x-32)

8 0

3 years ago

Find the inverse of the Function<br> <img src="https://tex.z-dn.net/?f=f%28x%29%3D%5Csqrt%5B3%5D%7B%5Cfrac%7Bx%7D%7B9%7D%20%7D%2

scoray [572]

9514 1404 393

Answer:

D

Step-by-step explanation:

Solve x = f(y) for y.

x = ∛(y/9) -4

x +4 = ∛(y/9) . . . . . add 4

(x +4)³ = y/9 .. . . . . cube

9(x +4)³ = y . . . . . . multiply by 9

The inverse function is ...

f⁻¹(x) = 9(x +4)³ . . . . . matches the last choice

6 0

3 years ago

If ABC is rotated 90° clockwise about the origin what will be the new coordinates of vertex B?

shutvik [7]

Answer:

B= (-1,-4)

Step-by-step explanation:

8 0

4 years ago

Intersection point of Y=logx and y=1/2log(x+1)

GalinKa [24]

Answer:

The intersection is $(\frac{1+\sqrt{5}}{2},\log(\frac{1+\sqrt{5}}{2})$ .

The Problem:

What is the intersection point of $y=\log(x)$ and $y=\frac{1}{2}\log(x+1)$ ?

Step-by-step explanation:

To find the intersection of $y=\log(x)$ and $y=\frac{1}{2}\log(x+1)$ , we will need to find when they have a common point; when their $x$ and $y$ are the same.

Let's start with setting the $y$ 's equal to find those $x$ 's for which the $y$ 's are the same.

$\log(x)=\frac{1}{2}\log(x+1)$

By power rule:

$\log(x)=\log((x+1)^\frac{1}{2})$

Since $\log(u)=\log(v)$ implies $u=v$ :

$x=(x+1)^\frac{1}{2}$

Squaring both sides to get rid of the fraction exponent:

$x^2=x+1$

This is a quadratic equation.

Subtract $(x+1)$ on both sides:

$x^2-(x+1)=0$

$x^2-x-1=0$

Comparing this to $ax^2+bx+c=0$ we see the following:

$a=1$

$b=-1$

$c=-1$

Let's plug them into the quadratic formula:

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$x=\frac{1 \pm \sqrt{(-1)^2-4(1)(-1)}}{2(1)}$

$x=\frac{1 \pm \sqrt{1+4}}{2}$

$x=\frac{1 \pm \sqrt{5}}{2}$

So we have the solutions to the quadratic equation are:

$x=\frac{1+\sqrt{5}}{2}$ or $x=\frac{1-\sqrt{5}}{2}$ .

The second solution definitely gives at least one of the logarithm equation problems.

Example: $\log(x)$ has problems when $x \le 0$ and so the second solution is a problem.

So the $x$ where the equations intersect is at $x=\frac{1+\sqrt{5}}{2}$ .

Let's find the $y$ -coordinate.

You may use either equation.

I choose $y=\log(x)$ .

$y=\log(\frac{1+\sqrt{5}}{2})$

The intersection is $(\frac{1+\sqrt{5}}{2},\log(\frac{1+\sqrt{5}}{2})$ .

6 0

3 years ago